Htykuut

I'm trying to evaluate $$displaystyle lim_{x to infty} x cdot left(dfrac{1}{x!}right)^{dfrac{1}{x}}$$

If you can't use the Stirling's approximation, how would you proceed?

Taking the logarithm,

$$-frac1nsum_{k=1}^nlogfrac kn$$ is a Riemannian sum.

$$-int_0^1log t,dt=left.t(log t-1)right|_0^1=1.$$

Hence $$e.$$

Let consider

$$a_n=n cdot left(dfrac{1}{n!}right)^{dfrac{1}{n}}=sqrt[n]{b_n}quad b_n=frac{n^n}{n!}$$

and by "ratio-root" criterion

$$frac{b_{n+1}}{b_n}=frac{(n+1)^{n+1}}{(n+1)!}frac{n!}{n^n}=left(1+frac1nright)^nto e implies a_n to e$$

Here a proof for the criterion without using $limsup$ and $liminf$ concept.

Let assume $frac{b_{n+1}}{b_n}to L$ then for any $epsilon >0$ exists $n_0in mathbb{N}$ such that

$$L-frac{epsilon}2le frac{b_{n+1}}{b_n} le L-frac{epsilon}2 quad forall nge n_0$$

then

$$left(L-frac{epsilon}2right)b_{n_0}le b_{n_0+1} leleft(L+frac{epsilon}2right)b_{n_0} $$

by induction we can easily show that

$$left(L-frac{epsilon}2right)^kb_{n_0}le b_{n_0+k} leleft(L+frac{epsilon}2right)^kb_{n_0} $$

$$left(L-frac{epsilon}2right)^nleft(L-frac{epsilon}2right)^{-n_0}b_{n_0}le b_{n} leleft(L+frac{epsilon}2right)^n left(L+frac{epsilon}2right)^{-n_0}b_{n_0} $$

and taking the $sqrt[n]{dots}$

$$left(L-frac{epsilon}2right)sqrt[n]{left(L-frac{epsilon}2right)^{-n_0}b_{n_0}}le sqrt[n]{b_{n}} leleft(L+frac{epsilon}2right)sqrt[n]{ left(L+frac{epsilon}2right)^{-n_0}b_{n_0}} $$

and then eventually

$$L-epsilon le sqrt[n]{b_{n}} le L+epsilon iff sqrt[n]{b_{n}}to L$$